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We need to find a criteria that is common to the specific columns where
your data is located. Actually, that should read: We need to find a criteria that is common and unique to the specific columns where your data is located. For example, if we tried using a divisor of 2 the remainder would be 0. But this criteria includes columns that are not the specific columns of interest. This criteria would include columns: 40,42,44,46,48,50,52 So, using a divisor of 6 sets the mod equal to 4 and this criteria is both common and unique to only those columns of interest. Biff "T. Valko" wrote in message ... Why does the MOD function get set to "=4"? We need to find a criteria that is common to the specific columns where your data is located. Column AN = column number 40 Column AT = column number 46 Column AZ = column number 52 This pattern of every 6th cell repeats to the end of your range. What is common about those columns is that when you divide the column number by 6 the remainder is 4: =MOD(COLUMN(AN1),6) = 4 =MOD(COLUMN(AT1),6) = 4 =MOD(COLUMN(AZ1),6) = 4 The formula limits the calculation to only those columns where the mod of the column number equals 4. Biff "SJT" wrote in message ... That's great. You have been incredibly helpful. One last thing. Why does the MOD function get set to "=4"? Promise not to bother you w/ this formula again. "T. Valko" wrote: These expressions will return arrays of either TRUE or FALSE (MOD(COLUMN(AN3:IV3),6)=4) (AN$1:IV$1<=TODAY()) The double unary "--" coerces those logical values to numeric values: --TRUE = 1 --FALSE = 0 See these for more info: http://mcgimpsey.com/excel/formulae/doubleneg.html http://xldynamic.com/source/xld.SUMPRODUCT.html Biff "SJT" wrote in message ... Thank you so much for all of your assistance. BTW what does the "--" before and after the MOD function do? Thanks again. Really appreciate it. "T. Valko" wrote: I interpreted the post differently than you. I read it to mean there is a single date for each month (the first of each month) spaced evenly every 6th cell starting from column AN and the OP is basically looking for a YTD sum. Biff "JLatham" <HelpFrom @ Jlathamsite.com.(removethis) wrote in message ... Damn, that's sweet - but take a look at it a little close, I'm coming up short 2 months. I filled each of the columns from 3 to 15 with increasing values starting at 1, so for 12 months with date after 3rd, I'd expect to get a total of 12 for that row, but I'm only getting 12. I put your formula into both columns AJ and AL with the same results: 10 instead of the expected 12. To test on out into the future I changed the formula to point to a cell with a manually entered date instead of using TODAY() just to test this kind of thing. For my test date I used 1/4/2008 and just cannot get it to come up with anything but 10 for those. But yours takes care of the whole situation much better than my ugly beastie in general - specifically the values in rows corresponding to days after current day of the month. "T. Valko" wrote: Try this: Row 1 = date headers =SUMPRODUCT(--(MOD(COLUMN(AN3:IV3),6)=4),--(AN$1:IV$1<=TODAY()),AN3:IV3) Adjust for the end of the range Copy down as needed Biff "SJT" wrote in message ... I think I'm in need of some type of index & match formula. I have column headings that are dates and below them are a series of numbers. I would like a formula that adds the numbers below the column headings based on today's date. For example, in column AN I have a column heading of 1/1/07 and below the column heading I have numerical data in rows 3 -15. In column AT I have the column heading 2/1/07 and in Column AZ, I have the column heading 3/1/07. This continues in the same pattern for each month of the year and below each column heading is data in rows 3 - 15. I would like to add the cumulative total of each row for each column depending on the date. Since today is 3/3, for example I would like to add the total of Cells AT3 and AN3 in one cell and then have a similar formula for each of the rows 4-15. On 4/1 this formula would add to the previous totals the data in Column AZ rows 3 - 15. Would appreciate any help you could provide. Thank you in advance for your assistance. |
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